Let $\mathbb{D}^2$ be the closed unit disk, and let $f:\mathbb{D}^2 \to \mathbb{C}$ be a smooth map, which is holomorphic on the open unit disk $\text{int}(\mathbb{D}^2)$.

Suppose that there exists a sequence $z_ n \in \text{int}(\mathbb{D}^2)$, $z_n \to z_0 \in \partial \mathbb{D}^2$ such that $f(z_n)=0$. Is $f$ identically zero on $ \mathbb{D}^2$?

The usual formulation of the identity theorem is for open connected domains; it states that a holomorphic function whose zero set has an accumulation point (inside the open domain) is identically zero.

Note that I assumed that $f$ is smooth on the closed disk. (In a sense it is "holomorphic" at the boundary too, as the condition of being conformal is a closed one).


If $f$ could be extend $f$ holomorphically to an open neighbourhood of $\mathbb D^2$, then the answer would be positive, by the usual identity theorem (as the accumulation point would now be in the interior of the new extended domain).

I am not sure if such an extension is always possible. There are certainly continuous examples that cannot be extended: e.g. $ f(z) = \sum_{n=1}^\infty \frac{z^{n!}}{n!}$. (See here for details). However, I don't know any smooth example which cannot be extended.

  • $\begingroup$ What exactly is your definition of "smooth" for a function on the closed disk that is different to the open disk? Smoothness is generally defined in terms of the derivatives, and derivatives require open regions to define. $\endgroup$ – or1426 May 1 '19 at 12:58
  • $\begingroup$ There is such a thing called "smooth maps between manifolds with boundary". In our present context, this means that around every boundary point $p \in \partial \mathbb D^2$ we can locally extend $f$ to a smooth function on an open neighbourhood. (Equivalently, all the partial derivatives of $f$ , of all orders, have continuous extensions up to the boundary). $\endgroup$ – Asaf Shachar May 1 '19 at 13:10
  • $\begingroup$ @or1426: In general, there is no holomorphic extension. For instance, take a bounded simply connected domain $U$ with $C^\infty$ boundary in the complex plane but the boundary is nowehere real analytic. Then use the Riemann mapping $f: D\to U$. This mapping has a $C^\infty$ extension to the boundary. This extended function has no holomorphic extension at any boundary point of the unit disk. mathoverflow.net/questions/82613/… $\endgroup$ – Moishe Kohan May 1 '19 at 15:17
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    $\begingroup$ I am not sure about the question itself but Luzin-Privalov theorem might suffice for your purposes of study of harmonic functions: If $f$ is zero at a subset of positive linear measure on the boundary of $D$ then $f$ is identically zero. I suggest, you post the question at MO and ping Alex Eremenko, I am sure he knows either a counterexample or a reference. $\endgroup$ – Moishe Kohan May 1 '19 at 15:43
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    $\begingroup$ On the second thought, no need to ask at MO, such functions doe exist and are not hard to find. See math.stackexchange.com/questions/74295/… for an example which is continuous along the boundary. One can use the same idea to find smooth examples. $\endgroup$ – Moishe Kohan May 1 '19 at 21:30

This is not a complete answer, but it was too long for a comment. As such, I am making it community wiki, if anyone wants to fill the missing steps.

An easy example (provided you know the Ostrowski-Hadamard gap theorem) of a $C^{\infty}$ map on the closed disk, holomorphic in the interior which cannot be extended (i.e. the circle is its natural boundary):

$$f(z)=\sum_{n=0}^{\infty} \frac{z^{2^n}}{n!}$$

Back to the original problem:

Consider the orizontal strip $\Omega=\{-\frac\pi2<\Im(z)<\frac\pi2\}$, and the conformal mapping $\varphi:\Omega\to \text{int}(\mathbb{D}^2)$, which is easily seen to be $\varphi(z)=\frac{e^z-1}{e^z+1}=\text{tanh}\left(\frac z2\right)$.

On the horizontal strip, we can consider the function $g(z):=\sin(z)h(z)$, where $h$ is a map that goes to $0$ suitably fast as $z\in \Omega\to \infty$. This function has infinite zeros, and thus precomposing with $\varphi^{-1}$ gives a holomorphic map on the unit disc with infinite zeros. It remains only to prove that it is smooth on the boundary: if $z\neq \pm 1$ this is trivial, and if we choose an $h$ that goes to $0$ fast enough, this should suffice to ensure smoothness. It surely is enough to ensure $n$-th differentiability


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